Question

This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 19.12 minutes and a standard deviation of 4.45 minutes.
Report your answers accurate to 2 decimals


a. What percent of runners took 19.4 minutes or less to complete the race? %


b. What time in minutes is the cutoff for the fastest 8.96 %? Minutes

Answer 1

a)

X ~ N ( µ = 19.12 , σ = 4.45 )
We convert this to standard normal as
P ( X < x ) = P ( Z < ( X - µ ) / σ )
P ( ( X < 19.4 ) = P ( Z < 19.4 - 19.12 ) / 4.45 )
= P ( Z < 0.06 )
P ( X < 19.4 ) = 0.5239 (From Z table)
= 52.39 %

b)

X ~ N ( µ = 19.12 , σ = 4.45 )
P ( X > x ) = 0.0896
​​​​​​ 1 - P ( X < x ) = 1 - 0.0896 = 0.9104
To find the value of x
Looking for the probability 0.9104 in standard normal table to calculate critical value Z = 1.3432
Z = ( X - µ ) / σ
1.3432 = ( X - 19.12 ) / 4.45
Solve for x
X = 25.10 min